Question

In Problems 37–72, solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent.

Short answer

The solution of the system is $x=-3,y=2,z=1$,or, using ordered triplets,$\left(-3,2,1\right)$ .

Step-by-step solution

Step 1. Given information. 

The given system of equation are

$\left\{\begin{array}{l}2x+y=-4\\ -2y+4z=0\\ 3x-2z=-11\end{array}\right.$

Step 2. Calculation. 

The augmented matrix of the system is:

$\left[\begin{array}{cc}\begin{array}{ccc}2& 1& 0\\ 0& -2& 4\\ 3& 0& -2\end{array}& \begin{array}{c}-4\\ 0\\ -11\end{array}\end{array}\right]$

Perform the row operations $R_3=2r_3-3r_2$:

role="math" $\left[\begin{array}{cc}\begin{array}{ccc}2& 1& 0\\ 0& -2& 4\\ 0& -3& -4\end{array}& \begin{array}{c}-4\\ 0\\ -10\end{array}\end{array}\right]$

Perform the row operations $R_3=2r_3-3r_2$:

$\left[\begin{array}{cc}\begin{array}{ccc}2& 1& 0\\ 0& -2& 4\\ 0& 0& -20\end{array}& \begin{array}{c}-4\\ 0\\ -20\end{array}\end{array}\right]$

Perform the row operations $R_2=-\frac{1}{2}{r}_2;R_3=-\frac{1}{20}{r}_3$:

$\left[\begin{array}{cc}\begin{array}{ccc}2& 1& 0\\ 0& 1& -2\\ 0& 0& 1\end{array}& \begin{array}{c}-4\\ 0\\ 1\end{array}\end{array}\right]$

Use the obtained matrix to write the system of equations.

$$\begin{gathered} 2x+y=-4 \\ y-2z=0 \\ z=1 \end{gathered}$$

Step 3. Solve the equation.  

Solve the equations to find the solution set.

Substitute $z=1$into the second equation.

$$\begin{gathered} y-2z=0 \\ y-2\left(1\right)=0 \\ y-2=0 \\ y=2 \end{gathered}$$

Substitute $y=2$into the first equation.

$$\begin{gathered} 2x+y=-4 \\ 2x+2=-4 \\ 2x=-6 \\ x=-\frac{6}{2}=-3 \end{gathered}$$

Therefore, the solution of the system is$x=-3,y=2,z=1$or, using ordered triplets,$\left(-3,2,1\right)$.

Step 4. Conclusion. 

Therefore, the solution of the system is $x=-3,y=2,z=1$or, using ordered triplets,$\left(-3,2,1\right)$.